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The question that continues to haunt me since I learned Wavefunctions and Schrodinger Eqns. is:

Why does, repulsion not occur due to orbital mixing since the dense electron cloud will still continue to repel the other pair.

The thought itself, comes from my understanding in the subject that, during opposite spin pairing, though repulsion is at max, but due to some conservative nature, it holds the electron density together.

But what actually relates to that conservative nature, and how come probabilities are temporarily created and broken when the bonds are broken.

How the hell does the same wavefunction (electron density) return back to same atom when the bond is broken?

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    $\begingroup$ The wave function always belongs to the entire system, i.e. when we are modeling an H2 molecule the wave function depends on six coordinates. Three for one electron and three for the other. In the Born-Oppenheimer approximation we are also changing the distance between the two protons, i.e. we are evaluating the electronic wave function for different distances, but we are treating the nuclei as classical charges. Even if we break the bond by making the distance between the nuclei infinite it stays one wavefunction, it just diagonalizes into two uncoupled hydrogen spectra. $\endgroup$ Commented Aug 1 at 4:37
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    $\begingroup$ @FlatterMann Ah! I got you . It clearly depends on our system consideration $\endgroup$
    – user418147
    Commented Aug 1 at 12:12
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    $\begingroup$ The number of quanta that are relevant to the system sets the dimensionality of the configuration space. It's really no different from classical mechanics, except that in classical mechanics positions and momenta are assumed to be independent. In quantum mechanics we have to chose if we want to be in a position or a momentum representation. For chemical problems we are mostly interested in spatial distributions for high energy scattering it's the distribution of momenta that we are seeking. $\endgroup$ Commented Aug 1 at 12:59

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